domain of f

**robinpatrick** · October 23rd 2006, 07:25 PM

f(x) = (square root -x) / x^3 + 2x^2 - 3x find the domain of f

**ThePerfectHacker** · October 23rd 2006, 07:33 PM

Originally Posted by robinpatrick

f(x) = (square root -x) / x^3 + 2x^2 - 3x find the domain of f

1)The square root needs to be non-negative.
Thus, $\sqrt{-x}$ for $-x\geq 0$ thus, $x\leq 0$ .

2)The numerator cannot be zero.
$x^3+2x^2-3x=0$
$x(x^2+2x-3)=0$
$x(x+3)(x-1)=0$
Thus,
$x\not = 0,1,-3$ .

Since $x\leq 0$ the only one of those that works is $x=-3$ thus,
$x\leq 0 \mbox{ and } x\not = -3$
Thus,
$(-\infty,-3)\cup (-3,0]$
-----
[Note for you, for anyone interested in anything I have to say].
These domain seem not be be well-defined, because to definied a function you need to create a domain and co-domain which you did not do here. Thus, how can you find a domain?

**topsquark** · October 24th 2006, 06:20 AM

Originally Posted by ThePerfectHacker

1)The square root needs to be non-negative.
Thus, $\sqrt{-x}$ for $-x\geq 0$ thus, $x\leq 0$ .

2)The numerator cannot be zero.
$x^3+2x^2-3x=0$
$x(x^2+2x-3)=0$
$x(x+3)(x-1)=0$
Thus,
$x\not = 0,1,-3$ .

Since $x\leq 0$ the only one of those that works is $x=-3$ thus,
$x\leq 0 \mbox{ and } x\not = -3$
Thus,
$(-\infty,-3)\cup (-3,0]$
-----
[Note for you, for anyone interested in anything I have to say].
These domain seem not be be well-defined, because to definied a function you need to create a domain and co-domain which you did not do here. Thus, how can you find a domain?

Slight correction. x cannot be 0 else the denominator is zero. So the domain would be:

$(-\infty,-3)\cup (-3,0)$

-Dan

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